Does adding a force on top of stressed cable, in its free portion, adds to the initial stress 4

[Grendizer]

A cable is stressed between two points, A on its first end, and B near its second end, by a force of 100KN, and a free portion near B remains without stress.

Please refer to attached image: Cable stressed before adding another force

Now B holds this 100KN by simply holding a lock attached on its cable. So if the cable is moved even by 1mm, this support vanishes.

Then the free end C is pushed with a force of 20KN. Now does force moves the cable away from its support at B, so the 100KN disappears? or does it adds to the 100KN to become a 120KN stress?

 

[Settingsun]

If I've understood you correctly, AB load remains 100kN; BC load is 20kN. Reaction at B is 80kN on the slab and 20kN by the pull force from C.

 

[robyengIT]

the reaction in B (on the support) is 80 kN but on the cable remain 100 kN

 

[BridgeSmith]

Steve and Roy are correct, IF the the strand remains locked off at B. However, you mention it moving. If the strand is allowed to slip through the anchor, the prestress force will be reduced according to the decrease in strain of the strand.

 

[Tommy385]

I believe that you will have that movement after your force C will be larger than 100. The prestress force will always be min. 100 (100+ force C) as long as the cable wont be "released".

 

[Ingenuity]

And this is the premise of 'couplings' (intermediate stressing) with post-tensioning tendons.

Neglecting friction (and other losses), at a magnitude of 100 kN in segment BC there will be "lift-off" at anchorage B, and uniform tension of 100 kN throughout A to C.

 

[Tommy385]

Ingenuity is right, and it seems that my english is getting rusty.

- if F2 is 20, A-B stays 100, B-C is 20 and anchor force is 80
- if F2 is 100, A-B stays 100, B-C is 100 and anchor force is 0
- if F2 is 120, A-B is 120, B-C is 120 and anchor force is 0

Of course neglecting friction (and other losses)....